Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 317 / 1

Work in an inertial frame centred on the centre of mass, and follow a fixed material mass, with no stellar wind or accretion through its boundary. Assume nonrelativistic motion, Newtonian self-gravity, no external gravitational field, and sufficiently regular integrable fields. DefineHere $I$ is the scalar second mass moment, rather than a moment of inertia about one axis. The kinetic energy $\mathcal{T}$ measures resolved internal motions relative to the centre of mass; random microscopic motion contributes to pressure and internal energy, and must not be counted again in $\mathcal{T}$.

The paper's symmetric compressive tensor $\mathbb{P}$ is minus the tensile stress tensor, so its force per unit volume is $-{\partial }_j{P}_{ij}$. Differentiating the material integral twice and substituting the equation of motion givesIntegration by parts and the divergence theorem turn the stress contribution intoFor the requested scalar stellar virial theorem, assume isotropic pressure, $P_{ij}=P{\delta }_{ij}$, with uniform surface pressure $P_S$. Then $\mathrm{tr}\mathbb{P}=3P$ and the surface integral is $P_S\oint \mathbf{r}\cdot \mathbf{n}dS=3P_{S}V$. No spherical shape is needed for this step.

Finally, symmetrizing the pair integral for the Newtonian gravitational potential givesThereforeAnisotropic magnetic or viscous stresses retain their full trace and boundary terms; a changing integration mass adds transport terms.

For an ideal gas with constant specific-heat ratio $\gamma$, the specific internal energy is $u=c_{V}T$ and $P=\rho (c_P-c_V)T=(\gamma -1)\rho u$. HenceThe stellar virial theorem consequently reads $\ddot{I}/2=2\mathcal{T}+3(\gamma -1)U-3P_{S}V+\Omega$. For a static nonrotating star with negligible surface pressure, $\mathcal{T}=\ddot{I}=P_S=0$, so $\Omega =-3(\gamma -1)U$.

Fully ionized nonrelativistic ions and electrons form a monatomic ideal gas, with $\gamma =5/3$, provided electron degeneracy pressure and radiation pressure are negligible. ThusBeing steady alone does not eliminate rotation or other macroscopic kinetic energy; those are additional assumptions needed for this relation.

Consider a spherical stellar polytrope with $P=K{\rho }^{1+1/n}$, constant $K$, $0<n<5$, zero surface pressure and finite radius $R$. Its specific enthalpy, measured from zero density, is $h={\int }_0^{P}d{P}^{\prime }/\rho =(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =\Phi (R)=-GM/R$. Integrating over mass therefore yieldsThe static stellar virial theorem gives $3\int PdV=-\Omega$, and eliminating the pressure integral proves the gravitational energy of a stellar polytrope:For an ideal gas with constant specific-heat ratio, the accompanying internal energy isTo express both energies using $\gamma$ alone one additionally identifies the model as an adiabatic stellar polytrope, so $n=1/(\gamma -1)$. ThenFor $\gamma =5/3$ these become $\Omega =-6G{M}^2/(7R)$ and $U=3G{M}^2/(7R)$. A general polytropic index need not be determined by the thermodynamic specific-heat ratio; without the adiabatic identification both $n$ and $\gamma$ remain.